# Engineering Mathematics Questions and Answers – Indeterminate Forms – 1

This set of Engineering Mathematics Multiple Choice Questions & Answers (MCQs) focuses on “Indeterminate Forms – 1”.

1. Find ltx → ∞$$(1+\frac{1}{x^2+2x+1})^{x^2+3x+1}$$
a) e
b) 1
c) e2
d) 1e

Explanation: Use the form
lt x→ ∞(1 + f(x))g(x) = elt x→ ∞ f(x) * g(x)
Provided as x → ∞ we must have
f(x) → 0
g(x) → ∞
These conditions are met in our question
L = ltx → ∞$$(1+\frac{1}{x^2+2x+1})^{x^2+3x+1} = e^{\frac{x^2+3x+1}{x^2+2x+1}}$$
ltx → 0$$\frac{x^2+3x+1}{x^2+2x+1}=1$$
L = e1 = e.

2. Find lt x → ∞$$(\frac{ln(1+\frac{(x+3)^3(2x+9)}{(4x^3+3)})}{x^3+3x^2+9x+27})$$
a) 0
b) 1
c) Undefined
d) – 135

Explanation: The form here is of 00
Applying L hospitals rule would be really tough to differentiate. Hence we use the concept of Taylor Series
We know that ln(1+x)=$$x-\frac{x^2}{2}+\frac{x^3}{3}-…\infty$$
Thus, we have
$$=lt_{x\rightarrow -3}(\frac{1}{x^3+3x^2+9x+27}\times (\frac{(x+3)^3(2x+9)}{4x^3+3}-\frac{(x+3)^6(2x+9)^2}{2(4x^3+3)^2}+..\infty))$$
$$=lt_{x\rightarrow -3}(\frac{1}{x+3}^3\times(\frac{(x+3)^3(2x+9)}{4x^3+3}-\frac{(x+3)^6(2x+9)^2}{2(4x^3+3)^2}+..\infty))$$
$$=lt_{x\rightarrow -3}(\frac{(2x+9)}{4x^3+3}-\frac{(x+3)^3(2x+9)^2}{2(4x^3+3)^2}+…\infty)$$
All the terms except the first one go to zero, we now have
$$=lt_{x\rightarrow -3}\frac{(2x+9)}{(4x^3+3)}=\frac{2(-3)+9}{4(-3)^3+3}=\frac{3}{-105}$$
$$=-\frac{1}{35}$$

3. Find ltn → ∞$$\sum_{a=0}^{n-1}\frac{sin(\frac{a}{n})}{n}$$
a) 1a
b) 1
c) 1 – cos(1)
d) 0

Explanation: We use the concept of limit of a sum which is
$$\int_a^b f(x)dx=lt_{n\rightarrow \infty}(\frac{b-a}{n})\times(f(a)+f(a+\frac{b-a}{n})….+f(a+\frac{(n-1)(b-a)}{n}))$$
Thus we have
$$\int_0^1 sin(x)dx=lt_{n\rightarrow \infty}(\frac{1}{n})\times(f(0)+f(\frac{1}{n})+….+f(\frac{n-1}{n}))$$
$$=lt_{n\rightarrow\infty}\frac{1}{n} \times (sin(\frac{1}{n})+sin(\frac{2}{n})+…+sin(\frac{n-1}{n}))$$
$$\int_0^1 sin(x)dx=lt_{n\rightarrow \infty}\sum_{a=0}^{n-1}\frac{sin(\frac{a}{n})}{n}$$
It is enough to evaluate the integral
$$\int_{0}^{1} sin(x)dx=[-cos(x)]_0^1$$=(cos(0)-cos(1))
=(1-cos(1))

4. Find ltx → 0$$(\frac{ln(1+x^4)}{x})$$
a) 1
b) -1
c) 0
d) Undefined

Explanation:
ltx → 0$$(\frac{ln(1+x^4)}{x}) = lt_{x\rightarrow 0} (\frac{1}{x}) \times (\frac{x^4}{1}-\frac{x^8}{2}+\frac{x^{12}}{3}-…\infty)$$
ltx → 0$$(\frac{x^3}{1}-\frac{x^7}{2}+\frac{x^{11}}{3}-…\infty)$$
= 0.

5. Find $$lt_{x\rightarrow 0}(\frac{1}{sin^2(x)})$$
a) 2
b) 1
c) 0
d) Undefined

Explanation:
=$$lt_{x\rightarrow 0}(\frac{1}{(\frac{1-cos(2x)}{2})})=lt_{x\rightarrow 0}(\frac{2}{1-cos(2x)})$$
=$$(\frac{2}{1-cos(0)})=\frac{2}{0} \rightarrow \infty$$
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6. Find $$lt_{x\rightarrow \infty}((\frac{x^3+x^2+x}{x^3+x+1})^{x+3})$$
a) e
b) e-1
c) 0
d) 1

Explanation:
=$$lt_{x\rightarrow\infty}(1+\frac{x^2-1}{x^3+x+1^{x+3}})^{x+3}=e^{lt_{x\rightarrow\infty}(\frac{(x^2-1)(x+3)}{x^3+x+1})}$$
$$lt_{x\rightarrow\infty}(\frac{(x^2-1)(x+3)}{x^3+x+1})=1$$
= e1 = e.

7. Find $$lt_{x\rightarrow -2}(\frac{sin((x-2)^2)}{(x+2)^2})$$
a) 1
b) 0
c) ∞
d) 00

Explanation: We have 00 form
Now we have the form
$$lt_{x\rightarrow a}\frac{sin(f(x))}{g(x)}$$=1
where f(x) → 0:g(x) → 0 as x → a
∴$$lt_{x\rightarrow -2}(\frac{sin((x-2)^2)}{(x-2)^2})=1$$

8. Find $$lt_{n\rightarrow\infty}\sum_{a=1}^{n-1}(\frac{ln(1+\frac{a}{n})}{n})$$
a) ln(2)
b) ln(4)
c) 3ln(2)
d) 1a

Explanation: Using limit of sum we have
=$$lt_{n\rightarrow\infty}(\frac{1}{n})\times(ln(1+0)+ln(1+\frac{1}{n})+ln(1+\frac{2}{n})+…+ln(1+\frac{(n-1)}{n}))$$
=$$\int_0^1 ln(1+x)dr=[(x+1)(ln(x+1)-1)]_0^1$$
=(2ln(2)-1)-(ln(1)-1)=2ln(2)
=ln(4)

9. Find $$lt_{n\rightarrow\infty}\frac{(1^a+2^a+…+(n-1)^a)}{n^{a+1}}$$
a) 1
b) 1a + 1
c) 0
d) Undefined

Explanation: $$\int_0^1x^a dx=lt_{n\rightarrow\infty}\frac{1}{n}\times((\frac{1}{n})^a+(\frac{2}{n})^a+…+(\frac{n-1}{n})^a)$$
=$$lt_{n\rightarrow\infty}\frac{(1^a+2^a+…+(n-1)^a)}{n^{a+1}}$$
It is enough to evaluatethis integral
$$\int_0^1 x^adx=[\frac{x^{a+1}}{a+1}]_0^1$$
=$$\frac{1}{a+1}$$

10. Find $$lt_{x\rightarrow -101}(\frac{ln(x^2+20x+(x+101)^2(x^2+3))-ln(x^2+20x)}{x^2+202x+10201})$$.
a) 0
b) 1
c) ∞
d) $$=\frac{10204}{12221}$$

Explanation: $$=lt_{x\rightarrow -101}(\frac{ln(1+\frac{(x+101)^2(x^2+3)}{(x^2+20x)}}{(x+101)^2})$$
Now expanding into Taylor series we have
$$=lt_{x\rightarrow -101}(\frac{1}{(x+101)^2})\times(\frac{(x+101)^2(x^2+3)}{(x^2+20x)}-\frac{(x+101)^4(x^2+3)^2}{2(x^2+20x)^2}+…\infty)$$
$$=lt_{x\rightarrow -101}(\frac{(x^2+3)}{(x^2+20x)}-\frac{(x+101)^2(x^2+3)^2}{2(x^2+20x)^2}+….\infty)$$
All others exceptthe first term tend to zero. Thus, we have
$$=lt_{x\rightarrow -101}(\frac{(-101)^2+3}{(101)^2+20(101)})$$
$$=\frac{10204}{12221}$$

11. Find $$lt_{x\rightarrow -5}(\frac{tan^{-1}(x^2+6x+5)}{x^2+15x+50})$$
a) 0
b) 1
c) -45
d) -1

Explanation: $$=lt_{x\rightarrow -5}\frac{tan^{-1}((x+5)(x+1))}{(x+5)(x+10)}$$
Now expand into Taylor Series for tan-1(x)
$$=lt_{x\rightarrow -5}\frac{1}{(x+5)(x+10)}\times((x+5)(x+1)-\frac{(x+5)^3(x+2)^3}{3}+\frac{(x+5)^5(x+2)^5}{5}-..\infty)$$
$$=lt_{x\rightarrow -5}(\frac{(x+1)}{(x+10)}-\frac{(x+5)^2(x+2)^3}{3(x+10)}+\frac{(x+5)^4(x+2)^5}{5(x+10)}-…\infty)$$
$$=lt_{x\rightarrow -5}\frac{(-5)+1}{(-5)+10}$$
=$$-\frac{4}{5}$$

12. Find $$lt_{x\rightarrow 0}\frac{sin((4x^3)tan^{-1}(x))}{x^4}$$
a) 1
b) 2
c) 4
d) 3

Explanation: Expand into Mclaurin series
$$=lt_{x\rightarrow 0}(\frac{1}{x^4})\times(\frac{4x^3 tan^{-1}(x)}{1!}-\frac{16x^6 tan^{-1}(x)}{3!}+…\infty)$$
$$=lt_{x\rightarrow 0}(\frac{4 tan^{-1}(x)}{x1!}-\frac{16x^2 tan^{-1}(x)}{3!}+…\infty)$$
Neglecting higher order terms (which go to zero) we have
$$=lt_{x\rightarrow 0}(\frac{4 tan^{-1}(x)}{x!})=lt_{x\rightarrow 0}(\frac{4}{x})\times(\frac{x}{1}-\frac{x^3}{3}+..\infty)$$
$$=lt_{x\rightarrow 0}(\frac{4 tan^{-1}(x)}{x!})=lt_{x\rightarrow 0}(\frac{4}{1}-\frac{4x^2}{3}+…\infty)=4$$

13. Find $$lt_{x\rightarrow 0}\frac{sin(sin(x))}{x}$$
a) 1
b) ∞
c) 0
d) -1

Explanation: $$=lt_{x\rightarrow 0}(\frac{1}{x})\times(\frac{sin(x)}{1!}-\frac{(sin(x))^3}{3!}+…\infty)$$
$$=lt_{x\rightarrow 0}(\frac{sin(x)}{x.1!}-\frac{(sin(x))^3}{x.3!}+…\infty)$$
$$=lt_{x\rightarrow 0}(\frac{sin(x)}{x}-lt_{x\rightarrow 0}\frac{sin(x)}{x.3!}\times(sin(x))^2+…\infty)$$
$$=lt_{x\rightarrow 0}\frac{sin(x)}{x}=1$$

14. Find $$=lt_{x\rightarrow 0}\frac{(a_nx^n+a_{n-1}x^{n-1}+…+a_1x+a_0)}{(b_nx^n+b_{n-1}x^{n-1}+…+b_1x+b_0)}$$
a) anbn
b) ∞
c) No general form
d) bnan

Explanation: $$=lt_{x\rightarrow 0}(\frac{x^n}{x^n})\times\frac{(a_n+a_{n-1}\frac{1}{x}+…+a_1\frac{1}{x^{n-1}}+a_0\frac{1}{x^{n}})}{(b_n+b_{n-1}\frac{1}{x}+…+b_1\frac{1}{x^{n-1}}+b_0\frac{1}{x^{n}})}$$
$$=\frac{a_n}{b_n}$$

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